Math 514 – Ch 1 Sep 18, 2018 Shawn Math 514 No comments yet Math 632 – 9/18 Sep 18, 2018 Shawn Math 632 1 comment Math 632 – 9/13 Sep 18, 2018 Shawn Math 632 No comments yet Math 632 – 9/11 Sep 18, 2018 Shawn Math 632 No comments yet Math 632 – 9/6 Sep 18, 2018 Shawn Math 632 No comments yet 1 … 8 9 10 11 12 … 60
![Motivation • Goal: given f(x)=0, find x • Motivation for numerical methods ○ ax+b=0⇒x=−b/a ○ ax^2+bx+c=0⇒x=(−b±√(b^2−4ac))/2a ○ ⋮ ○ ax^5+bx^4+cx^3+dx^2+ex+f=0⇒ No formula! • If the order of polynomial is ≥5, there is no explicit zero formula Theorem 1.1: Bolzanos theorem • Statement ○ Let f be a real-valued continuous function on the interval [a,b] ○ If f(a)f(b)≤0, then ∃ξ∈[a,b] s.t. f(ξ)=0 • Explanation ○ If a continuous function has values of opposite sign inside an interval ○ Then it has a root in that interval • Proof ○ By the Intermediate Value Theorem • Note ○ This theorem does not guarantee the uniqueness of solution Theorem 1.2: Brouwer’s Fixed Point Theorem • Statement ○ If g∈C, and g(x)∈[a,b] for x∈[a,b] ○ Then ∃ξ∈[a,b] s.t. g(ξ)=ξ ○ Here, the real number ξ is called the fixed point of g • Proof ○ Let f(x)≔x−g(x) ○ Then, f(a)f(b)=⏟((a−g(a)) )┬(≤0) ⏟((b−g(b)) )┬(≥0)≤0 ○ By Intermediate Value Theorem, ∃ξ∈[a,b] s.t. f(ξ)=0 ○ ⇒f(ξ)=ξ−g(ξ)=0 ○ ⇒g(ξ)=ξ • Why care about fixed point? ○ Finding fixed point is numerically easier in the sense of iteration • Algorithm ○ Initial guess: x_0∈[a,b] ○ Iterate: x_(k+1)≔g(x_k ) ○ Stop when |x_(n+1)−x_n |<ε, where ε is a small number • Example ○ Given g(x)=1/2 (x^2+1/2) ○ The fixed point of g should satisfy x=1/2 (x^2+1/2)⇔x^2−2x+1/2=0 ○ Let f(x)≔x^2−2x+1/2, then we need to find f(x)=0 ○ Analytical method § x=1±√2/2≈1.7 or 0.3 ○ Numerical method § x_0=1 § x_1=g(x_0 )=g(1)=3/4=0.75 § x_2=g(x_1 )=g(3/4)=17/32≈0.53 § x_3=g(x_2 )=g(17/32)≈0.39 § ⋮ • Counter-example ○ Suppose f(x)=x^2−2 ○ f(x)=0⇔x^2=2⇔x=2/x ○ Let x_(k+1)=2/x_k and x_0=1, then x_1=2, x_2=1,x_3=2⋯ Two Main Questions Over This Chapter • When does x_(k+1)=g(x_k ) converge? ○ If the iteration is unstable § x_(k+1)=g(x_k ) diverges ○ If the iteration is stable § The contraction argument guarantees convergence § And the convergence rate is linear • Given f(x), how to find g(x)? ○ There are infinitely many g for a given f as long as f(x)=0⇔g(x)=x ○ Possible choice for g(x) § g(x)=x+f(x), or § g(x)=x+ln(f(x)+1) ○ Newtons method (and secant method) will guarantee a contracting g(x) Theorem 1.3 & 1.4 & 1.5: Contraction Mapping Theorem • Contraction ○ Let g be a real-valued continuous function on the interval [a,b] ○ Then g is a contraction on [a,b] if ∃L∈(0,1) s.t. ○ |g(x)−g(y)|≤L|x−y|,∀x,y∈[a,b] (Lipschitz condition) ○ Here, L is called Lipschitz constant • Remark on Lipschitz condition ○ |g(x)−g(y)|≤L|x−y|, ∀x,y∈[a,b] ○ ⇒|g(x)−g(y)|/|x−y| ≤L ○ ⇒lim┬(y→x)〖|g(x)−g(y)|/|x−y| 〗≤L ○ ⇒|g^′ (x)|≤L<1 (assume g is differentiable) • Statements ○ Let g be a contraction on [a,b] ○ Suppose g(x)∈[a,b], ∀x∈[a,b]. Then, (1) ∃ξ∈[a,b] s.t. g(ξ)=ξ (i.e. There exists a fixed point) (2) {x_(k+1)=g(x_k )} converges to ξ, ∀x_0∈[a,b] (i.e. The iterative algorithm works) (3) If the iteration stop at |x_k−ξ|≤ε, then k≤1+[ln〖|x_1−x_0 |−ln(ε(1−L)) 〗/ln(1/L) ] where [x] is the largest integer less than or equal to x • Proof for (1) ○ See Theorem 1.2 • Proof for (2) ○ ⏟(|x_(k+1)−ξ| )┬(E_(k+1) )=|g(x_k )−g(ξ)| ○ ≤L ⏟(|x_k−ξ| )┬(E_k ) by Lipschitz condition ○ ≤L^2 ⏟(|x_(k−1)−ξ| )┬(E_(k−1) ), since ⏟(|x_k−ξ| )┬(E_k )≤L ⏟(|x_(k−1)−ξ| )┬(E_(k−1) ) by induction ○ ⋮ ○ ≤L^(k+1) ⏟(|x_0−ξ| )┬(E_0 )→0 as k→∞ • Proof for (3) ○ From the proof for (2), we know that § E_k≤L^k E_0≤ε ○ Taking log on both side, we obtain § k≤log_L〖ε/E_0 〗 ○ Calculate E_0 § E_0=|x_0−ξ|=|x_0−x_1+x_1−ξ| ≤|x_0−x_1 |+|x_1−ξ|≤|x_0−x_1 |+L|x_0−ξ| § ⇒E_0≤〖|x_0−x_1 |+L⋅E〗_0 § ⇒E_0≤|x_1−x_0 |/(1−L) ○ Therefore § k≥log_L〖ε/(|x_1−x_0 |/(1−L))〗=log_L〖ε(1−L)/|x_1−x_0 | 〗=ln〖|x_1−x_0 |−ln(ε(1−L)) 〗/ln(1/L) • Corollary ○ Given g:[a,b]→[a,b], and g∈C^1 [a,b] ○ If |g^′ (x)|≤L<1, then the sequence {x_k=g(x_(k−1) )} converges to ξ • Remark on Corollary ○ If we relax |g^′ (x)|<1 to be just |g′(ξ)|<1 ○ Then when x_0 is close to ξ, {x_k } will converge to ξ ○ Since in a small neighborhood of ξ, g^′ (x)~|g^′ (ξ)|<1 Theorem 1.3: Stability of Fixed Point • Stable Fixed Point ○ If ξ=g(ξ), and |g^′ (ξ)|<1, then ξ is a stable fixed point ○ A stable fixed point can be found via {x_(k+1)=g(x_k )} • Unstable Fixed Point ○ If ξ=g(ξ), and |g^′ (ξ)|>1, then ξ is a unstable fixed point ○ If ξ is an unstable fixed point, then {x_(k+1)=g(x_k )} wont converge to ξ Definition 1.4 & 1.7: Rate of Convergence • Suppose ξ=lim┬(k→∞)〖x_k 〗, and define E_k≔|x_k−ξ| • An algorithm is said to converge linearly if ○ lim┬(k→∞)〖E_(k+1)/(E_k^ )〗=μ, for some constant μ∈(0,1) • An algorithm is said to converge superlinearly if ○ lim┬(k→∞)〖E_(k+1)/E_k 〗=0 • An algorithm is said to converge quadratically if ○ lim┬(k→∞)〖E_(k+1)/(E_k^2 )〗=μ, for some constant μ>0 • An algorithm is said to converge with order q if ○ lim┬(k→∞)〖E_(k+1)/(E_k^q )〗=μ, for some constant μ>0 Example 1.7: f(x)=e^x−x−2 • g(x)=e^x−2 ○ We observed that g(x) maps [1,2] to [1,2] § By the Fixed Point Theorem, ∃ξ∈[1,2] s.t. g(ξ)=ξ § We need to check whether g(x) satisfies the Lipschitz condition § g^′ (ξ)=e^ξ∈[e^1,e^2 ] § ⇒|g^′ (ξ)|>1 § ⇒ unstable fixed point § ⇒ the algorithm wont work ○ And g(x) also maps [−2,−1] to [−2,−1] § By the Fixed Point Theorem, ∃ξ∈[1,2] s.t. g(ξ)=ξ § g^′ (ξ)=e^ξ∈[e^(−2),e^(−1) ] § ⇒|g^′ (ξ)|<1 § ⇒ stable fixed point § ⇒ run {x_(k+1)=g(x_k )} for ξ • g(x)=ln(x+2) ○ We observed that g(x) maps [1,2] to [1,2] § g^′ (ξ)=1/(ξ+2)∈[1/4,1/3] § ⇒|g^′ (ξ)|<1 § ⇒ stable fixed point § ⇒ run {x_(k+1)=g(x_k )} for ξ ○ And g(x) also maps (−2,−1) to (−2,−1) § g^′ (ξ)=1/(ξ+2)∈(1,+∞) § ⇒|g^′ (ξ)|>1 § ⇒ unstable fixed point § ⇒ the algorithm wont work • Remark ○ x=e^x−2⇒f(x)=e^x−x−2 § We have a stable fixed point ξ∈[−2,−1], and a unstable ξ∈[1,2] ○ x=ln(x+2)⇒e^x=x+2⇒f(x)=e^x−x−2 § We have a stable fixed point ξ∈[1,2], and a unstable ξ∈[−2,−1] ○ Therefore the choice of g will affect the convergence behavior ○ So how can we design a function g(x) s.t. every fixed point is stable? Definition 1.6: Newton’s Method • In Newtons method, g(x) is defined as ○ g(x)=x−f(x)/(f^′ (x) ) • Its obvious that f(ξ)=0⇔g(ξ)=ξ • Why the fixed points of g is stable ○ We want to show that |g^′ (ξ)|<1 ○ g(x)=x−f(x)/(f^′ (x) )⇒g^′ (x)=1−(f(x)/(f^′ (x) ))^′ ○ (f/f^′ )^′=(f^′⋅f^′−f⋅f′′)/(f^′ )^2 =1−(f⋅f′′)/(f^′ )^2 ○ ⇒g^′ (x)=1−(1−(f(x)⋅f^′′ (x))/[f^′ (x)]^2 )=(f(x)⋅f^′′ (x))/[f^′ (x)]^2 ○ ⇒|g^′ (ξ)|=(f(ξ)⋅f^′′ (ξ))/[f^′ (ξ)]^2 =0<1 Theorem 1.8: Convergence of Newtons Method • Statement ○ In Newtons method, {x_(k+1)=g(x_k )} converges quadratically ○ i.e. lim┬(k→∞)〖E_(k+1)/(E_k^2 )〗≤μ<1 • Assumption ○ f(ξ)=0 ○ f∈C^2 in [ξ−δ,ξ+δ]=I_δ, since we need to use f^′ and f^′′ ○ f^′ (ξ)≠0, since it will appear at the denominator ○ |(f^′′ (x))/(f^′ (y) )|≤A, (∀x,y∈I_δ ) ○ |x_0−ξ|≤1/A (i.e. The initial guess is not too far away from ξ) • Proof ○ Expand f(ξ) at x_k to obtain f(x_k ) and f′(x_k ) § f(ξ)=f(x_k+ξ−x_k ) § =f(x_k )+f^′ (x_k )(ξ−x_k )+1/2 f^′′ (x_k ) (x_k−ξ)^2+… (by Taylor expansion of f) § =f(x_k )+f^′ (x_k )(ξ−x_k )+1/2 f^′′ (θ_k ) (x_k−ξ)^2 (for some constant θ_k∈(x_k,ξ), by the Mean Value Theorem) ○ Express f(x_k )/f′(x_k ) in x_(k+1) using f^′′/f′, since we already know |(f^′′ (x))/(f^′ (y) )|≤A § By assumption, f(ξ)=0 § ⇒f(x_k )+f^′ (x_k )(ξ−x_k )+1/2 f^′′ (θ_k ) (x_k−ξ)^2=0 § ⇒f(x_k )=−1/2 f^′′ (θ_k ) (x_k−ξ)^2−f^′ (x_k )(ξ−x_k ) § ⇒f(x_k )/f′(x_k ) =−1/2 (f^′′ (θ_k ))/(f^′ (x_k ) ) (ξ−x_k )^2−(ξ−x_k ) ○ Compute E_(k+1), and express it with E_k § E_(k+1)=|x_(k+1)−ξ|=|g(x_k )−ξ| § =|(x_k−f(x_k )/(f^′ (x_k ) ))−ξ| § =|{x_k−[−1/2 (f^′′ (θ_k ))/(f^′ (x_k ) ) (ξ−x_k )^2−(ξ−x_k )]}−ξ| § =|x_k+1/2 (f^′′ (θ_k ))/(f^′ (x_k ) ) (ξ−x_k )^2+ξ−x_k−ξ| § =1/2 |(f^′′ (θ_k ))/(f^′ (x_k ) )| (ξ−x_k )^2 § =1/2 |(f^′′ (θ_k ))/(f^′ (x_k ) )| E_k^2 ○ Show the algorithm converges § By assumption,|x_k−ξ|≤1/A, and |(f^′′ (x))/(f^′ (y) )|≤A, (∀x,y∈I_δ ) § So, E_(k+1)=1/2 ⏟(|(f^′′ (θ_k ))/(f^′ (x_k ) )| )┬(≤A) ⏟(E_k^ )┬(≤1/A) E_k^ ≤1/2⋅A⋅1/A⋅E_k=1/2 E_k→0 as k→+∞ § Therefore x_k converges to ξ ○ Show the algorithm converges quadratically § E_(k+1)/(E_k^2 )=1/2 |(f^′′ (θ_k ))/(f^′ (x_k ) )|≤1/2 A § As k→+∞, both x_k and θ_k converge to ξ § Thus, lim┬(k→+∞)〖E_(k+1)/(E_k^2 )〗=1/2 |(f^′′ (ξ))/(f^′ (ξ) )|=μ, where μ∈(0,A/2] is a constant Definition 1.8: Secant Method • Motivation ○ Sometimes f^′ can be hard to find in Newtons method ○ But we can approximate f′ using a difference quotient ○ i.e. f^′ (x_k )≈(f(x_k )−f(x_(k−1) ))/(x_k−x_(k−1) ) • Definition ○ x_(k+1)=x_k−f(x_k )∕((f(x_k )−f(x_(k−1) ))/(x_k−x_(k−1) )) =x_k−f(x_k )((x_k−x_(k−1))/(f(x_k )−f(x_(k−1) ) )) • Note ○ The secant method requires two initial values x_0 and x_1 Theorem 1.10: Convergence of Secant Method • Statement ○ Let f∈C^1 [ξ−δ,ξ+δ] s.t. f(ξ)=0 and f^′ (ξ)≠0 ○ If x_0,x_1 is close to ξ, then {x_(k+1)=g(x_k )} converges at least linearly • Proof ○ WLOG, assume α≔f^′ (ξ)>0 in a small neighborhood of ξ ○ Choose I be a neighborhood of ξ such that § 0<3/4 α<f^′ (x)<5/4 α,∀x∈I ○ Compute x_(k+1) § x_(k+1)=x_k−f(x_k )∕((f(x_k )−f(x_(k−1) ))/(x_k−x_(k−1) )) § By the Mean Value Theorem □ (f(x_k )−⏞(f(ξ) )┴0)/(x_k−ξ)=f^′ (η_k )⇒f(x_k )=f^′ (η_k )(x_k−ξ),and □ (f(x_k )−f(x_(k−1) ))/(x_k−x_(k−1) )=f^′ (θ_k ) □ for some θ_k∈[x_k,x_(k−1) ],η_k∈(x_k,ξ) § Therefore, x_(k+1)=x_k−(f^′ (η_k )(x_k−ξ))/(f^′ (θ_k ) ) ○ Check E_(k+1)/E_k <1 § E_(k+1)=x_(k+1)−ξ=E_k−(f^′ (η_k ))/(f^′ (θ_k ) ) E_k=[1−(f^′ (η_k ))/(f^′ (θ_k ) )] E_k § ⇒E_(k+1)/E_k =[1−(f^′ (η_k ))/(f^′ (θ_k ) )]<(1−(5α∕4)/(3α∕4))=2/3<1 ○ Therefore secant method converges at least linearly](https://shawnzhong.com/wp-content/uploads/2018/09/img_5ba1653853572.png)
